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Let $(X,\mathcal{M},\mu)$ be a measure space. Take $\epsilon > 0$ and $A \in \mathcal{M}$.

Is it in general true that there exist and open set $U$ such that $A \subseteq U$ and $\mu(U \setminus A)<\epsilon$ ?

If it is, I would also be grateful for a hint on how to prove this.

I need only the case when $X=\mathbb{R^n}$, $\mathcal{M} = \mathcal{B}(\mathbb{R^n})$, and $\mu$ is the Lebesgue measure. But I would be interested in a more general result, if there is one.

Thank you in advance!

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up vote 5 down vote accepted

On an arbitrary set $X$, the concept of "open set" is not defined. In order to talk about open sets, you need $X$ to be a topological space.

A very common general setting is the following: Let $X$ be a locally compact Hausdorff space, $\mathcal{M} \supset \mathcal{B}(X)$ a $\sigma$-algebra containing the Borel sets, and $\mu$ a Borel measure on $\mathcal{M}$. Your question amounts to asking whether or not the measure $\mu$ is "outer regular." In general, no, an arbitrary Borel measure on $X$ need not be outer regular.

However, it is true that the Lebesgue measure on $\mathbb{R}^n$ is outer regular. Here's a hint:

If you're defining Lebesgue measure as $$\mu(A) = \inf\{\sum_{n=1}^\infty |I_n|\colon\, A \subset \bigcup_{n=1}^\infty I_n\},$$ where the $I_n$ are open intervals (rectangles), then you should think about using the definition of "infimum."

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Thank you, exactly the kind of answer I wanted! :) –  DoubleTrouble Nov 14 '12 at 12:26
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